I get that f(x) = P_n(x) + R(x), and calculating P_n(x) is no big deal--it's just a finite Taylor series. But I am struggling with the R(x). Case in point is a function such as f(x) = xe^(x^2). P_n(x) is just x^3 + x. But what the heck do I do with the R(x)? I know it's a big-O of x^4, since that's the next term, but what on earth do I do from there?

Sep 22nd 2012, 11:45 PM

johnsomeone

Re: Cannot figure out the remainder function of a Taylor series

The point is usually to *bound* the remainder term. That will typically involve using the specifics of the domain on which the approximation is to hold.

If is the degree n Taylor polynomial about for ( is times differentiable), then

, where for some between and .

----

In your example, if , then to find the remainder after the 3rd degree Taylor poly at do:

.

.

.

.

Thus , giving , just as you said.

Now to bound the remainder term, we need to specify the domain on which we're using this approximation.

Let's suppose we want to approximate by pn the interval , for some . Note .

APOLOGIES - that's terrible notation. I'm using capital R for both the real number defining the interval, and the Taylor remainder term. I hope the context makes it clear which is which. I don't feel like trying to go back and change it everywhere.

Then for some between and , where .

And thus for some .

But then .

So we want to bound where .

, so

, so

.

Thus .

Suppose . Then .

Suppose . Then

Suppose . Then

so that

It's customary to keep those x powers there, but you could bound it absolutely on the specified domain by inserting , in which case:

If . Then .

If . Then .

If . Then .

Obviously, the approximation can get real bad real fast (not surprising given how explosive is), but if we're sticking to the interval (-1/2, 1/2), approximating by is a pretty good approxiation.

There's of course a fair amount of algebraic skill in figuring the upper bounds for the remainder term. But that should hopefully give you some idea.